Vector spaces and dimension: unordered pairs Let $K= \mathbb{Z}_p$, where $p$ is a prime number, and let V be a vector space over the field K such that $\dim{V} = 3$.

I have no idea where to start with this, I'm not even really sure what I'm being asked. Can anyone point me in the right direction?
 A: i. Take a basis {a,b,c} of V over K. Then $V = \{k_1a + k_2b + k_3c \ | \ k_1,k_2,k_3 \in K\}$, and so V has $p^3$ elements (one for each choice of $k_1,k_2,k_3)$, because K has $p$ elements. 
For an unordered pair of vectors to be linearly independent, they must be both non-zero. You can pair every non-zero vector $v$ with any non-zero vector distinct from $v,2v,...,(p-1)v$, because those are exactly the non-zero vectors linearly dependent with $v$. Therefore the number of unordered pairs of linearly independent vectors is: \begin{equation} \frac{1}{2}(p^3-1)\cdot[(p^3-1)- (p-1)]= \frac{1}{2} (p^3-1)(p^3-p). \end{equation}
ii. Given a pair $(v,w)$ of linealry independent vectors, the pairs of vectors that span the same dimension 2 subspace are those of the form $(av+bw,cv+dw)$ that are linearly independent. We count them: We can choose $a$ and $b$ of $p^2-1$ different ways (because they can't be both zero), and we can choose $c$ and $d$ of $p^2-1$ different ways minus the $p-1$ ways that result in multiples of $av+bw$. We divide by 2 since we do not care of the order of pairs. Therefore, the number of unordered pairs spanning the same space as $(v,w)$ is:
\begin{equation}
 \frac{1}{2}(p^2-1)\cdot[(p^2-1)-(p-1)]=\frac{1}{2}(p^2-1)(p-1)p.
\end{equation}
Now, to get the number of distinct subspaces of dimension 2, note that each of them is generated by an unordered pair of linearly independent vectors: we have $\frac{1}{2} (p^3-1)(p^3-p)$ of them. But for each one of them, $\frac{1}{2}(p^2-1)(p-1)p$ give the same space. Therefore, there are
\begin{equation}
\frac{\frac{1}{2} (p^3-1)(p^3-p)}{\frac{1}{2}(p^2-1)(p-1)p} = p^2+p+1
\end{equation}
disctinct subspaces of dimension 2.
A: I'll add two assumptions here: $V$ is a vector space over $\mathbb Z_p$, and $p$ is a prime natural number.
Since $V$ is a $3$-dimensional vector space over $\mathbb Z_p$, we can say that $V$ is the same as (at least isomorphic to) the set of all ordered triples $(a_1,a_2,a_3)$ where each $a_i$ is in $\mathbb Z_p$. We know that $\mathbb Z_p$ has exactly $p$ elements, so how many ordered triples are possible? That answers the first question in part $1$.
Two vectors in $V$ are linearly independent if neither is a multiple of the other. Remember that any non-zero vector in $V$ has $p$ multiples, and if the first vector is a non-zero multiple of the second, then the second is also a multiple of the first. So you can count the linearly dependent pairs by taking two groups: where one vector is zero, and where both are non-zero and they are multiples of each other. This should make the count fairly easy.
Is this enough of a pointer?
